# Julia set

Let $U$ be an open subset of the complex plane and let $f:U\to U$
be analytic. Denote the $n$-th iterate of $f$ by ${f}^{n}$, i.e. ${f}^{1}=f$
and ${f}^{n+1}=f\circ {f}^{n}$. Then the *Julia set* of $f$ is the
subset $J$ of $U$ characterized by the following property: if $z\in J$
then the restriction of $\{{f}^{n}\mid n\in \mathbb{N}\}$ to any neighborhood^{}
of $z$ is not a normal family.

It can also be shown that the Julia set of $f$ is the closure of the set of repelling periodic points of $f$. (Repelling periodic point means that, for some $n$, we have ${f}^{n}(z)=z$ and $|{f}^{\prime}(z)|>1$.)

A simple example is afforded by the map $f(z)={z}^{2}$; in this case, the Julia set is the unit circle. In general, however, things are much more complicated and the Julia set is a fractal.

From the definition, it follows that the Julia set is closed under $f$ and its inverse — $f(J)=J$ and ${f}^{-1}(J)=J$. Topologically, Julia sets are perfect and have empty interior.

Title | Julia set |
---|---|

Canonical name | JuliaSet |

Date of creation | 2013-03-22 17:15:26 |

Last modified on | 2013-03-22 17:15:26 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 7 |

Author | rspuzio (6075) |

Entry type | Definition |

Classification | msc 28A80 |